Past exam of the classical mechanics course of Part B of the physics course of the University of Oxford / 2016 / 1

In a frame rotating with angular velocity $\mathbf{\Omega }$, a particle of relative velocity $\mathbf{v}$ experiences the Coriolis forceFor nearly horizontal atmospheric motion, the relevant vertical rotation component is $\Omega \sin \ell$, with latitude $\ell$. In the northern hemisphere the horizontal deflection is to the right of the motion. A low-pressure center produces an inward pressure-gradient force; an anticlockwise circulation has an outward Coriolis force to oppose it. A high-pressure center instead supports clockwise circulation, with inward Coriolis force opposing the outward pressure-gradient force.

The sense reverses in the southern hemisphere. There, anticlockwise circulation is anticyclonic and corresponds to high atmospheric pressure. HenceThis is the large-scale rotating-fluid interpretation; friction, centrifugal acceleration and departures from ideal balance modify actual wind profiles. NASA's description independently identifies the southern, counterclockwise Great Red Spot as a high-pressure system.^[ref]

Let $\mathsf{R}(\alpha )$ denote the planar rotation matrix, and choose the initial cup center on the positive $x$ axis, as in the requested equations. Its trajectory is $\mathbf{C}(t)=R(\cos \Omega t,\sin \Omega t)$. The handle rotates relative to the radial line at signed rate $\varphi$, so its absolute angular velocity is $\nu =\Omega +\varphi$. A positive recurrence time is $2\pi /|\varphi |$; the sign records rotation direction.

Writing $\mathbf{q}=(x_1,y_1)$, the relation isor explicitlyOne obtains the same result by rotating cup coordinates through $\varphi t$ into the radially aligned axes, then rotating through $\Omega t$ into the inertial axes.

At $\varphi =-\Omega$, $\nu =0$, andThe axes maintain a fixed orientation in inertial space while their origin accelerates around a circle. This is translation without rotation of the cup frame; it is not an inertial frame because the origin accelerates. In particular its inertial correction is a spatially uniform fictitious force, rather than a Coriolis force from rotation of these axes.

Treat the marble as a frictionless point particle; no size, rolling constraint or friction law is supplied. Neglect its vertical kinetic energy as requested, but retain the gravitational potential energy $mgz=m{\omega }^2(x_1^2+y_1^2)/2$.

For completeness keep the general constant cup rotation rate $\nu =\Omega +\varphi$. In cup components the center's velocity is $\mathbf{w}=R\Omega (\sin \varphi t,\cos \varphi t)$. With $\mathsf{J}(x,y)=(-y,x)$, the horizontal velocity norm is $|\dot{\mathbf{q}}+\nu \mathsf{J}\mathbf{q}+\mathbf{w}|$. ThusIn the nonrotating translating case $\varphi =-\Omega$, this reduces toThe time-only constant $m{R}^2{\Omega }^2/2$ may be dropped. Integrating the cross-velocity term by parts gives an equivalent LagrangianIt differs by a total derivative and a time-only term, so gives the same trajectories but different expressions for the canonical momenta. The small-vertical-velocity approximation is typically consistent at small surface slope ${\omega }^2|\mathbf{q}|/g$; it need not remain valid during unlimited resonant growth.

For the original Lagrangian of solution, differentiation with respect to the generalized velocities givesIn contrast, using $L^{\prime }$ in the nonrotating case gives $p_{x_1}^{\prime }=m{\dot{x}}_1$, $p_{y_1}^{\prime }=m{\dot{y}}_1$; this is the expected momentum shift under adding a total derivative.

The Euler-Lagrange equations in the general cup frame areThese include the Coriolis and centrifugal terms when the cup axes really rotate. Set $z_q=x_1+i{y}_1$ to obtainFor ${\omega }^2\ne {\Omega }^2$, its complete solution iswhere the two complex constants provide the four real initial data. The driving denominator is independent of cup-axis rotation: rotation changes the coordinate description, not the physical trap frequency.

In the requested special case $\nu =0$, an equivalent real form isFor example $a_x=x_1(0)-D$, $b_x={\dot{x}}_1(0)/\omega$, $a_y=y_1(0)$ and $b_y=[{\dot{y}}_1(0)-D\Omega ]/\omega$ determine the constants.

At resonance ${\omega }^2={\Omega }^2$ with nonzero $\Omega$, replace the particular term byIn the nonrotating cup frame this has components $(R\Omega t\sin \Omega t/2,-R\Omega t\cos \Omega t/2)$, plus the homogeneous solution. Secular growth eventually invalidates a finite cup or the neglected vertical kinetic energy. If the ride does not rotate, its forcing vanishes and ordinary free oscillations remain.